Recent content by PhMichael

I obviously did that. If I assume that the ball does not slip during impact then this means that its normal velocity vanishes, which is not true because it leaves the surface. Am I getting something wrong here?

The sphere is released at a height H above a fixed inclined plane, as shown in the attached figure. The coefficient of restitution at impact is e>0 (that is the sphere leaves the surface just after impact), the coefficient of friction between the sphere and the plane is \mu. I need a...

Homework Statement The attached images shows everything. ({\bf{e}}_1 denotes the direction of X and {\bf{e}}_2 denotes the direction of Y). Initially, the spring is force free when X_{0}=0.4 m (which yields Y_{0}=0.3 m). Also, at this instant, the velocity of B is...

I don't understand your point. You can't assume that the kinetic energy of the particle at A is zero in order to solve for the minimum value of h to reach that point because you'll get nothing out of this. Similarly, you shouldn't assume that the kinetic energy of the table is zero when its CM...

Well, this problem looks to me very similar to the particle problem which I presented previously and thus the logic should be the same. If not, then why?

Yeah, I meant a "linear impulse" and that's why I indicated it as an impulsive force. But what do you think about my problem? Isn't what I had written correct?

\hat{F}=\int_{t_1}^{t_2}{Fdt} \thinspace,\thinspace t_{2} \to t_{1}

I think that this problem is very similar to the classical problem which is attached in the figure below. Here, you should find the min. value of h for which the particle would reach the point A, which is above the center of the hoop C, and pass it. In order to do that, you apply the...

I'm arguing with a friend regarding the condition for a table to flip over assuming that the right contact point of its right leg with the ground (denoted hereafter by O) is held fixed i.e. the table cannot slip. The upper horizontal plate is a sort of a rod with mass m and both legs are...

The direction of the force from the spring should be "+" (to the right). However, I'm not sure what the answer of your first question. The spring is compressed by an amount l_{0} - w with respect to the box, however, I need to express it with respect to the fixed left end and thus, this...

-k\cdot{(x-b\cdot{sin(\omega \cdot t)}-l_{0})}=-m\cdot b \cdot \omega^{2} sin(\omega \cdot t) \Rightarrow \boxed{x = \frac {m}{k} \cdot b \cdot \omega^{2} \cdot sin(\omega \cdot t) + b\cdot sin(\omega \cdot t)+l_{0}} Now, \frac {m}{k} \cdot b \cdot \omega^{2} \cdot sin(\omega \cdot t) + b\cdot...

Homework Statement Well, I have this spring (stiffness k and free length l_{0}) mass ( m ) system in a box which has a width w such that l_{0}>w (i.e. the spring is compressed). The box is excited (given a prescribed position) by: u(t)=b\cdot sin({\omega}\cdot t). Determine the range of...

http://img831.imageshack.us/img831/1715/plotb.png Uploaded with ImageShack.us I need to plot Re(\lambda) as a function of \beta in order to find the stability range. According to the figure, the system is stable for \beta > 0.4457 , while for \beta<0.4457, there exists at least one...

Thanks a lot, viscousflow! ... After an intense struggle, I've managed to write a working code =) ... It goes like this: G=0.8896; beta=0:0.01:1; A=zeros(6,size(beta)); a1 = 1 ; a2 = 6.*beta - G ; a3 = 8.*beta.^2 - 4.*beta.*G + 6 ; a4 = beta.^3 - G.*beta.^2 + 16.*beta - 4.*G ...

After analyzing a 3-DOF system, I've obtained the following 6th order characteristic polynomial: P=\lambda^6+(6\beta-.8896)\lambda^5+(8\beta^2-3.5584\beta+6)\lambda^4+(\beta^3-.8896\beta^2+16\beta-3.5584)\lambda^3+(3\beta^2-1.7792\beta+8)\lambda^2+(3\beta-.8896)\lambda+1 Stability is...